Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/246



Test the series

$$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$$

Solution. Here

$$u_n = \frac{1}{\left ( n - 1 \right ) ! }, \; u_{n+1} = \frac{1}{n!} $$

$$\therefore \lim_{n \to \infty}\left( \frac{u_{n+1}}{u_n} \right) = lim_{n \to \infty}\left( \frac{\frac{1}{n}}{\frac{1}{\left(n-1\right)!}}\right) = \lim_{n \to \infty}\left( \frac{\left( n - 1 \right)!}{n!}\right) = \lim_{n \to \infty}\left( \frac{1}{n} \right) = 0 \left( = p \right)$$

and by I, §141, the series is convergent.

Test the series

$$\frac{1!}{10} + \frac{2!}{10^2} + \frac{3!}{10^3} + \cdots $$

Solution. Here

$$u_n = \frac{n!}{10^n}, \; u_{n+1} = \frac{\left( n + 1 \right)!}{10^{n+1}}$$

$$\therefore \lim_{n \to \infty} \left(\frac{u_{n+1}}{u_n}\right) = \lim_{n \to \infty}\left( \frac{\left( n + 1 \right)!}{10^{n+1}} \times \frac{10^n}{n!} \right) = \lim_{n \to \infty}\left( \frac{n+1}{10} \right) = \infty \left( = p \right)$$

and by II,§141 the series is divergent.

Test the series

Solution. Here

$$u_n = \frac{1}{\left( 2n - 1 \right) 2n}, \; u_{n+1} \frac{1}{\left( 2n + 1 \right)\left( 2n + 2 \right)}$$

$$\therefore \lim_{n \to \infty}\left( \frac{u_{n+1}}{u_n}\right) = \lim_{n \to \infty}\left[ \frac{\left( 2n - 1 \right)2n}{\left( 2n + 1 \right)\left( 2n + 2 \right)}\right] = \frac{\infty}{\infty}$$

This being an indeterminate form, we evaluate it, using the rule in §112.

Differentiating,

$$\lim_{n \to \infty}\left( \frac{8n - 2}{8n - 6} \right) = \frac{\infty}{\infty}$$

Differentiating again,

$$ \lim_{n \to \infty}\left( \frac{8}{8} \right) = 1 \left( = p \right)$$

This gives no test (III, §141). But if we compare series (C) with (G), §138,making $$p = 2$$, namely,

we see that (C) must be convergent, since its terms are less than the corresponding terms of (D), which was proved convergent.